Write down the cubic equation given that $\alpha + \beta + \gamma = 4$, $\alpha^2 + \beta^2 + \gamma^2 = 66$, and $\alpha^3 + \beta^3 + \gamma^3 = 280$
Ok so, the sum of roots is given and I'm able to use the sum of the roots and the sum of the roots squared to get the sum of the combination of roots, but I'm unable to get the product of roots, because I can't seem to manipulate the sum of the cubes of roots to resemble the sum of roots and sum of squares of roots.
Best Answer
$$\alpha\beta+\alpha\gamma+\beta\gamma=\frac{(\alpha+\beta+\gamma)^2-\alpha^2-\beta^2-\gamma^2}{2}=\frac{16-66}{2}=-25.$$ $$\alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)+3\alpha\beta\gamma.$$ Thus, $$280=64-3\cdot4\cdot(-25)+3\alpha\beta\gamma,$$ which gives $$\alpha\beta\gamma=-28$$ and we got the following equation. $$x^3-4x^2-25x+28=0.$$